# 2006 AMC 10B Problems/Problem 15

## Problem

Rhombus $ABCD$ is similar to rhombus $BFDE$. The area of rhombus $ABCD$ is $24$ and $\angle BAD = 60^\circ$. What is the area of rhombus $BFDE$?

$[asy] defaultpen(linewidth(0.7)+fontsize(10)); size(120); pair A=origin, B=(2,0), C=(3, sqrt(3)), D=(1, sqrt(3)), E=(1, 1/sqrt(3)), F=(2, 2/sqrt(3)); pair point=(3/2, sqrt(3)/2); draw(B--C--D--A--B--F--D--E--B); label("A", A, dir(point--A)); label("B", B, dir(point--B)); label("C", C, dir(point--C)); label("D", D, dir(point--D)); label("E", E, dir(point--E)); label("F", F, dir(point--F)); [/asy]$

$\textbf{(A) } 6\qquad \textbf{(B) } 4\sqrt{3}\qquad \textbf{(C) } 8\qquad \textbf{(D) } 9\qquad \textbf{(E) } 6\sqrt{3}$

## Solution 1

Using the property that opposite angles are equal in a rhombus, $\angle DAB = \angle DCB = 60 ^\circ$ and $\angle ADC = \angle ABC = 120 ^\circ$. It is easy to see that rhombus $ABCD$ is made up of equilateral triangles $DAB$ and $DCB$. Let the lengths of the sides of rhombus $ABCD$ be $s$.

The longer diagonal of rhombus $BFDE$ is $BD$. Since $BD$ is a side of an equilateral triangle with a side length of $s$, $BD = s$. The longer diagonal of rhombus $ABCD$ is $AC$. Since $AC$ is twice the length of an altitude of of an equilateral triangle with a side length of $s$, $AC = 2 \cdot \frac{s\sqrt{3}}{2} = s\sqrt{3}$.

The ratio of the longer diagonal of rhombus $BFDE$ to rhombus $ABCD$ is $\frac{s}{s\sqrt{3}} = \frac{\sqrt{3}}{3}$. Therefore, the ratio of the area of rhombus $BFDE$ to rhombus $ABCD$ is $\left( \frac{\sqrt{3}}{3} \right) ^2 = \frac{1}{3}$.

Let $x$ be the area of rhombus $BFDE$. Then $\frac{x}{24} = \frac{1}{3}$, so $x = \boxed{\textbf{(C) }8}$.

## Solution 2

Triangle DAB is equilateral so triangles $DEA$, $AEB$, $BED$, $BFD$, $BFC$ and $CFD$ are all congruent with angles $30^\circ$, $30^\circ$ and $120^\circ$ from which it follows that rhombus $BFDE$ has one third the area of rhombus $ABCD$ i.e. $8 \Longrightarrow \boxed{\textbf{(C) }8}$.

Note: A quick way to visualize this method is to draw the line $DB$ and notice the two equilateral triangles $\triangle ADB$ and $\triangle DBC$.

## See Also

 2006 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byProblem 16 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions

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